Suppose that $G$ is a reductive group defined over a field $k$ which is not quasisplit. Suppose that $S$ is a maximal $k$-split torus. Let $\mathcal{L}(S)$ be the centraliser of $S$ and $\mathcal{DL}(S)$ the commutator group of $\mathcal{L}(S)$ or the anisotropic kernel. Is it correct that $\mathcal{L}(S)=\mathcal{DL}(S)\cdot S$?

$\begingroup$No. Assume $G$ is semisimple (else silly counterexamples via non-split central tori). For a maximal $k$-torus $T\supset S$ and basis $\Delta$ of the set $\Phi$ of absolute roots, a basis of absolute roots of $Z_G(S)$ is $\Delta_0=\ker(\Delta\rightarrow{\rm{X}}(S))$. You want $\Delta_0$ of size $\dim(T)-\dim(S)$; i.e., $D:=\Delta -\Delta_0$ of size $\dim(S)$ $(\dim(T)=\#\Delta$ since $G$ is semisimple). Restriction $D\rightarrow{\rm{X}}(S)$ maps onto a basis of relative roots, so you want it injective; i.e., $\#D$ is the $k$-rank. False for many non-quasi-split unitary groups over fields.$\endgroup$
– user76758Dec 3 '13 at 14:50

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$\begingroup$I think there are also some E$_6$ counterexamples as well (over suitable fields). For "most" types it is OK by inspection of classification theorems.$\endgroup$
– user76758Dec 3 '13 at 14:53

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The question needs a little more context, which I'll try to fill in. The comments by user76758 seem to answer the question correctly in the negative, but if I have time later I'll try to add a concrete counterexample based on the Tits classification (summarized by him in lecture ntoes for the 1965 AMS summer institute at Boulder and published in Proc. Symp. Pure Math. 9). In that classification, relevant examples require one to start with a semisimple group $G$ of rank at least 3, but one also has to work over a suitable number field or such. Over a finite field $k$ all $G$ are quasisplit, and here one has $T = Z_G(S)$ in Borel-Tits notation, with $T$ a maximal $k$-torus containing a maximal $k$-split torus $S$.

As Tits indicates for a general connected reductive group $G$ over $k$, one has to distinguish between the semisimple anisotropic kernel $H:=(Z_G(S), Z_G(S))$ (derived group) and the reductive anisotropic kernel $Z_a \cdot H$ (almost-direct product), where $Z_a \subset T$ is the $k$-anisotropic part of the center $Z$ of $H$. As in the quasisplit case it's sometimes possible for $Z_a$ to be nontrivial in other cases, hence for $S$ to be properly contained in $Z$.

This is all part of the 1965 Borel-Tits structure theory (translated later into scheme language) for reductive groups over arbitrary fields. Here $Z_G(S)$ is a Levi subgroup of a minimal $k$-parabolic subgroup, etc. And of course the classification depends heavily on the nature of $k$, being usually incomplete when $G$ is $k$-anisotropic.

ADDED: From a quick look at the Tits classification list for simple types, it seems that counterexamples to the question asked occur (as user76758 suggested) in various cases involving outer automorphisms of the Dynkin diagrams. There are some in types $^2\!A_n$ and $^2\!D_n$ which appear over the real numbers as well as number fields, along with a couple in type $^2\!E_6$. In the latter case, Tits remarks that one of them occurs for a real form "which gives rise to a bounded symmetric domain". In his notation, $r$ is the relative rank ($=\dim S$), while $n$ is the absolute rank. Here $r$ is the number of Galois orbits in the Dynkin diagram, so you want $r$ to be strictly less than the number of simple roots in those orbits.

Concerning the Tits survey article, details (and corrections to the proof of the main theorem) were written down in a 1976 Bonn thesis by his student Martin Selbach (published in the series Bonner Math. Schriften).

$\begingroup$Thanks. Could I ask something a bit different. Let's say I had an element h in the semisimple anisotropic kernel such that its conjugacy class in the semisimple anistropic kernel is of positive dimension. Can I say that this conjugacy class is equal to the conjugacy class of $h$ in $Z_{G}(S)$?$\endgroup$
– RupertDec 4 '13 at 10:03

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$\begingroup$@Rupert: For any connected reductive group $H$ (such as $Z_G(S)$) and any $h \in \mathscr{D}(H)$, we know its conjugacy class in $H$ is the same as its conjugacy class in $\mathscr{D}(H)$. In what sense is your question is not an instance of that?$\endgroup$
– user76758Dec 4 '13 at 12:33

$\begingroup$@Eupert: Yes, the "absolute" theory pointed out by user76758 is clear, and applies to rational points over $k$ (here all semisimple, with classes of dim $>0$ unless central). But notation gets confusing at times (different uses of $H$ for instance).$\endgroup$
– Jim HumphreysDec 4 '13 at 13:52